The differential equation corresponding to primitive $y = e^{d x}$ is

or

The elimination of the arbitrary constant m from the equation $y = e^{m x}$ gives the differential equation

[MP PET 1995, 2000; Pb. CET 2000]

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C

$y = e^{m x} \Rightarrow l o g y m x \Rightarrow m = \frac{l o g y}{x} y = e^{m x} \Rightarrow \frac{d y}{d x} = m e^{m x} = \frac{l o g y}{x} y = (\frac{y}{x}) l o g y .$

Suggest Corrections

Similar questions

The differential equation corresponding to primitive $y = e^{d x}$ is

The elimination of the arbitrary constant m from the equation $y = e^{m x}$ gives the differential equation

[MP PET 1995, 2000; Pb. CET 2000]

The integrating factor of the differential equation $\frac{d y}{d x} = y t a n x - y^{2} s e c x$ is

[MP PET 1995; Pb. CET 2002]

y + x \frac{d y}{d x} = x - y \frac{d y}{d x}

If $y = c e^{s i n^{- 1} x},$ then corresponding to this the differential equation is

(y - x \frac{d y}{d x} - a (y^{2} + \frac{d y}{d x})

Join BYJU'S Learning Program